June 25, 2024, 02:27:28 AM
Forum Rules: Read This Before Posting


« previous next »
Pages: [1]   Go Down

Topic: Percents to Kg?  (Read 4164 times)

0 Members and 1 Guest are viewing this topic.

Offline Hemidol

  • Regular Member
  • ***
  • Posts: 68
  • Mole Snacks: +2/-2
  • Rune of Punishment
Percents to Kg?
« on: September 13, 2009, 06:56:56 PM »
An ore contains 16.0 % of the mineral bismuthinite, Bi2S3, which is a source of the element Bi. How
much ore must be processed in order to obtain 12.0 kg of Bi?


That's another problem I'm having trouble with...


Would I need to find the molar mass of Bi2S3 to solve this question? Basically I'm thinking find the molar mass, take 16% of that, (units will be in grams) and then I'm lost...
Logged

Offline MrTeo

  • Chemist
  • Full Member
  • *
  • Posts: 312
  • Mole Snacks: +31/-9
  • Gender: Male
Re: Percents to Kg?
« Reply #1 on: September 14, 2009, 01:27:20 AM »
You need to find the molar mass, but the 16% doesn't refer to bismuthinite, it's the ore's percentage of the mineral. That's why you also have to calculate the percentage of Bi in Bi2S3 (using atomic masses) and then apply these information to find the ore needed.
Logged
The way of the superior man may be compared to what takes place in traveling, when to go to a distance we must first traverse the space that is near, and in ascending a height, when we must begin from the lower ground. (Confucius)

Offline Hemidol

  • Regular Member
  • ***
  • Posts: 68
  • Mole Snacks: +2/-2
  • Rune of Punishment
Re: Percents to Kg?
« Reply #2 on: September 14, 2009, 07:03:49 PM »
Quote from: MrTeo on September 14, 2009, 01:27:20 AM
You need to find the molar mass, but the 16% doesn't refer to bismuthinite, it's the ore's percentage of the mineral. That's why you also have to calculate the percentage of Bi in Bi2S3 (using atomic masses) and then apply these information to find the ore needed.
Alright so this is what I did....

I figured in 100g of ore there's 16.0% (this is where I'm confused!!) and found the molar mass of Bi2S3 to be :

209.0 + 209.0 + 32.07 + 32.07 + 32.07 = 514.21 g / mole and therefore I say 1 mole contains 418 g of Bi (or is it just 209.0g?) and I set up my equation as such....


514.21 g / mol Bi2S3 * 100g ore / 16.0g Bi2S3

= 51421 / 16.0
= 3213.8125


. Is this correct? I think I did something wrong, if I did could you show me the correct method? :)

Thanks for the reply!
Logged

Offline DrCMS

  • Chemist
  • Sr. Member
  • *
  • Posts: 1298
  • Mole Snacks: +211/-81
  • Gender: Male
Re: Percents to Kg?
« Reply #3 on: September 15, 2009, 06:34:55 AM »
You really think that 100g or ore contains 3213.8125g of Bi?

You have correctly calculated the molecular mass of Bi2S3, well pretty close I get 514.17g with Bi 208.98g.

But from then on your calculations make no sense at all.  Think again.

What % Bi in Bi2S3?

What %Bi in an ore that is 16% Bi2S3?

How much ore to get 12.0kg of Bi?
Logged

Offline Hemidol

  • Regular Member
  • ***
  • Posts: 68
  • Mole Snacks: +2/-2
  • Rune of Punishment
Re: Percents to Kg?
« Reply #4 on: September 17, 2009, 04:00:59 PM »
Here I redid it, what do you think of this:


(100g ore / 17g Bi2S3) * (514.21g Bi2S3 / 1 mole Bi2O3) * (1 mole Bi2S3 / 2 mole Bi) * (1 mole Bi / 209g Bi) * (1000g Bi / 1Kg Bi) * (12Kg Bi) = 86835 ore should be processed
Logged

Offline DrCMS

  • Chemist
  • Sr. Member
  • *
  • Posts: 1298
  • Mole Snacks: +211/-81
  • Gender: Male
Re: Percents to Kg?
« Reply #5 on: September 17, 2009, 04:32:17 PM »
If you use the correct 16% ore content rather than 17% and put units on the answer you'll get the right answer. 

I'd also put the answer in Kg as the amount of Bi in the question is in Kg.
Logged
Pages: [1]   Go Up
« previous next »
Sponsored Links